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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    The length of second's pendulum is 1 m on earth. If mass and diameter of a planet is doubled than that of earth, then its length becomes:

    A) 1 m                                        

    B) 2 m

    C) 0.5 m     

    D) 4 m

    Correct Answer: C

    Solution :

    \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[2=2\pi \sqrt{\frac{{{R}^{2}}l}{GM}}\]                 ??..(i) Here,\[l=1m\]) and on given planet, \[2=2\pi \sqrt{\frac{R{{'}^{2}}l'}{GM'}}\] \[\Rightarrow \]               \[2=2\pi \sqrt{\frac{4{{R}^{2}}l'}{G\times 2M}}\]                               ?..(ii) From Eqs. (i) and (ii), \[2l'=l\] \[\Rightarrow \]               \[l'=1/2=0.5\,m\]                


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